Month: January 2023

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202301101150 Solution to 1999-IBHL-PHY-I-7

Two 10\,\mathrm{kg} blocks on a smooth horizontal surface are tied together. They are accelerated by a horizontal force of 30\,\mathrm{N} which acts as shown below:

If frictional effects are negligible, what is the tension in the connecting rope?

Roughwork.

As always start with free-body diagrams. Take rightward positive. Considering first the bodies as a whole:

so as to write the equation of motion by Newton 2nd law:

\begin{aligned} F_{\textrm{net}} & =ma \\ 30 & = 20a\\ a & = 1.5\,\mathrm{m\,s^{-2}} \\ \end{aligned}

where the tension regarded as internal force does not count. Considering then as individuals,

so as to write:

\begin{aligned} F_{\textrm{net}} & =ma \\ 30 - T & = (10)(1.5)\\ T & = 15\,\mathrm{N} \\ \end{aligned}

One may also do with the left block, but an exercise left the reader.

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202301041207 Solution to 1969-HL-PHY-2

Derive the relation between Young’s modulus Y and Hooke’s constant k for a uniform steel wire of cross-section A and length L.

Two uniform wires of equal cross-section have the same Young’s modulus Y but different Hooke’s constants k_1 and k_2. If the two wires are connected in series to form a new wire, compare Young’s modulus of the connected wire with that of the original wires, and find Hooke’s constant of the connected wire in terms of k_1 and k_2.

Roughwork.

\displaystyle{\textrm{Young's modulus (}Y\textrm{)}=\frac{\textrm{Stress (}\sigma\textrm{)}}{\textrm{Strain (}\varepsilon\textrm{)}}}

Y \equiv \displaystyle{\frac{\sigma}{\varepsilon} = \frac{F/A}{\Delta l/L} = \frac{FL}{A\Delta l}\quad \textrm{\scriptsize{OR}}\quad \displaystyle{F = \frac{YA\Delta l}{L}}

Lemma.

Hooke’s law for a stretched wire:

\displaystyle{F=\bigg(\frac{YA}{L}\bigg)\Delta l=kx}

such that

\displaystyle{k\equiv \frac{YA}{L}}\qquad\because\enspace x\equiv \Delta l.

Wikipedia on Young’s modulus

\begin{aligned} \Delta l' & = \Delta l_1+\Delta l_2 \\ \frac{F}{A'}\frac{L'}{Y'} & = \frac{F}{A}\bigg(\frac{L_1}{Y_1}+\frac{L_2}{Y_2}\bigg) \\ \because &\enspace \begin{cases} A' = A \\ Y_1 =Y_2 =Y\\ L' =L_1+L_2 \\ \end{cases} \\ \therefore& \quad Y'=Y \\ \end{aligned}

Hence

\begin{aligned} k' & = \frac{Y'A'}{L'} \\ & = \frac{YA}{L_1+L_2} \\ & = \bigg(\frac{L_1}{YA}+\frac{L_2}{YA}\bigg)^{-1} \\ & = \bigg(\frac{1}{k_1}+\frac{1}{k_2}\bigg)^{-1} \\ & = \bigg(\frac{k_1+k_2}{k_1k_2}\bigg)^{-1} \\ & = \frac{k_1k_2}{k_1+k_2} \\ \end{aligned}

This problem is not to be attempted.